The Direct Comparison Test (not to be confused with the Limit Comparison Test) says, paraphrased:
Choose some series #suma_n# and some other series #sumb_n# such that #a_n >= 0# and #b_n >= 0#, while #a_n <= b_n#. When that is done, the following limit is said to be a valid test:
If #sumb_n# is convergent, then #suma_n# is as well.
If #suma_n# is divergent, then #sumb_n# is as well.
Notice how the larger series determines convergence, not the smaller one. It's exactly like this:
If b, then a.
If not a, then not b.
which is a statement and then its contrapositive, which must be true. The affirmative is the convergence, and the denial is the divergence.
This is not true:
If b, then a.
If a, then b.
(If #a_n# converges, that says nothing about #b_n#.)
Now that that's out of the way, let's pick #a_n# and #b_n#. Picking #a_n# is easy; it's your series. #b_n#, however, must be always greater than #a_n#, and also have a lower bound at #0# just like #b_n#.
An easy way to pick #b_n# is to let #b_n# be #a_n# with the far-right constant smaller, since the denominator will now be smaller and the overall would be larger.
#suma_n = sum_(n=1)^(oo) 1/(5n^2 + 5)#
#sumb_n = sum_(n=1)^(oo) 1/(5n^2 + 4)#
Notice that since we're going towards #oo#, the constants don't even matter by then.
Now we need to show whether #b_n# is convergent, or whether #a_n# is divergent. As a form of a p-series (#sum 1/(n^p + k)# converges when #p > 1#), there's no doubt it's going to converge, but let's just perform the sum to see how it goes.
#1/(5(1)^2 + 4) + 1/(5(2)^2 + 4) + 1/(5(3)^2 + 4) + 1/(5(4)^2 + 4)#
# = 1/9 + 1/24 + 1/49 + 1/84 + ...#
#= 0.bar11 + 0.041bar6 + 0.0204 + 0.0119 + ...#
The denominator is approaching 0 pretty quickly in the first few values of #n#, and will continue rather quickly at each successive #n# (#n^2# gets very steep pretty quickly). So therefore, #a_n# must converge as well.
Wolfram Alpha says the convergence result is #pi/10cothpi - pi/10#, so there is indeed a finite sum for #a_n#. The sum for #b_n# is similar in magnitude but you can type that into Wolfram Alpha and see the exact answer if you're curious.
